1
$\begingroup$

The series $\sum_{n = 1}^{\infty} a_{n}$ with $a_{n} = \frac{1}{\sqrt n - i n}$ does not converge absolutely because, $$|a_{n}| = \frac{1}{|\sqrt n - i n|} = \frac{1}{\sqrt{n + n^2}} \geq \frac{1}{\sqrt 2 \ n } $$ and $\frac{1}{\sqrt 2}\sum_{n=1}^{\infty} \frac{1}{n}$ does not converge.

Therefore the ratio test and other comparison tests dont work. My question is, if this series does converge or if it diverges. I dont managed to show if the series is either a cauchy sequence or not

$\endgroup$
  • $\begingroup$ You are correct. It does not converge absolutely. In fact, it does not converge at all. $\endgroup$ – GEdgar Jun 24 '18 at 21:02
  • $\begingroup$ Im sure there are converging series that dont converge absolutely $\endgroup$ – ensisun Jun 24 '18 at 21:07
  • $\begingroup$ $\sum\frac{(-1)^n}{n}$ converges, but not absolutely. $\endgroup$ – GEdgar Jun 24 '18 at 21:09
2
$\begingroup$

$$\frac{1}{\sqrt{n}-in} = \frac{\sqrt{n}}{n+n^2}+i\frac{n}{n+n^2} $$ From the definition of a complex series, it doesn't converge, since it's imaginary part doesn't converge. ($\frac{n}{n+n^2} $ is comparable with $\frac{1}{n}$ and $\sum \frac{1}{n} $ doesn't converge)

$\endgroup$
  • $\begingroup$ When the absolute value of a complex number (in your case of the denominator of the summand) comes out negative, you can be pretty sure that you have made a mistake. $\endgroup$ – Alex B. Jun 24 '18 at 21:11
  • $\begingroup$ @AlexB. Thank you $\endgroup$ – Jakobian Jun 24 '18 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.